CODEWITHSUNDEEP
pythonpermutationpython-itertools
Neon Flash
Neon Flash

Reputation: 3233

How can I generate Permutations for multiple groupings in a list in Python

I know that we can use itertools.permutations to permutate different items in a list however, what if I have a list such that few items need to be in fixed positions, few items need to be swapped with one more and few items need to be swapped with 2 more?

For example:

test = [1, 6, 2, 12, 5, 13, 11, 14, 15]

How can I use Python itertools.permutation or another method to generate all possible combinations with the following constraints?

Update:

1 and 5 have fixed positions
In position 2, I could have either 6 or 11
In position 3, I could have either 2 or 12
In position 4, I could have 2 or 12
In position 6, I could have either 13, 14, 15 and so on

So, my list looks like this:

[1, (6, 11), (2, 12), (2,12), 5, (13, 14, 15), (6, 11), (13, 14, 15), (13, 14, 15)]

I have included the numbers in groups which represents that numbers in the same group can be swapped with each other.

Thanks.

Upvotes: 2

Views: 1525

Answers (1)

Dani Mesejo
Dani Mesejo

Reputation: 61930

You could do something like this:

from itertools import permutations, product, chain

test = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]
groups = [[1], [2, 3, 4], [5], [6, 7], [8, 9], [10], [11, 12, 13], [14], [15, 16]]

result = [list(chain.from_iterable(permutation)) for permutation in product(*map(permutations, groups))]

for e in result[:20]:
    print(e)

Output

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 12, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 12, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 11, 13, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 11, 13, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 11, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 11, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 11, 12, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 11, 12, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 12, 11, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 12, 11, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 11, 12, 13, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 11, 12, 13, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 11, 13, 12, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 11, 13, 12, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 12, 11, 13, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 12, 11, 13, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 12, 13, 11, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 12, 13, 11, 14, 16, 15]

UPDATE

Given the new constraints you can do something like this:

from functools import partial
from itertools import combinations, permutations, product, chain

choose_one = partial(lambda r, iterable: combinations(iterable, r), 1)
groups = [[[1]], combinations([6, 11], 1), permutations([2, 12]), [[5]], combinations([13, 14, 15], 1)]

for e in product(*groups, repeat=1):
    print(list(chain.from_iterable(e)))

Output

[1, 6, 2, 12, 5, 13]
[1, 6, 2, 12, 5, 14]
[1, 6, 2, 12, 5, 15]
[1, 6, 12, 2, 5, 13]
[1, 6, 12, 2, 5, 14]
[1, 6, 12, 2, 5, 15]
[1, 11, 2, 12, 5, 13]
[1, 11, 2, 12, 5, 14]
[1, 11, 2, 12, 5, 15]
[1, 11, 12, 2, 5, 13]
[1, 11, 12, 2, 5, 14]
[1, 11, 12, 2, 5, 15]

Upvotes: 3

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